Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case.

Describe the relative growth rates of functions.

In this section, we examine a powerful tool for evaluating limits. This tool, known as
L’Hôpital’s rule , uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value.

Applying l’hôpital’s rule

L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider

limx→af(x)g(x).

If
limx→af(x)=L1andlimx→ag(x)=L2≠0, then

limx→af(x)g(x)=L1L2.

However, what happens if
limx→af(x)=0 and
limx→ag(x)=0? We call this one of the
indeterminate forms , of type
00. This is considered an indeterminate form because we cannot determine the exact behavior of
f(x)g(x) as
x→a without further analysis. We have seen examples of this earlier in the text. For example, consider

limx→2x2−4x−2andlimx→0sinxx.

For the first of these examples, we can evaluate the limit by factoring the numerator and writing

limx→2x2−4x−2=limx→2(x+2)(x−2)x−2=limx→2(x+2)=2+2=4.

For
limx→0sinxx we were able to show, using a geometric argument, that

limx→0sinxx=1.

Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.

The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions
f and
g such that
limx→af(x)=0=limx→ag(x) and such that
g′(a)≠0 For
x near
a, we can write

f(x)≈f(a)+f′(a)(x−a)

and

g(x)≈g(a)+g′(a)(x−a).

Therefore,

f(x)g(x)≈f(a)+f′(a)(x−a)g(a)+g′(a)(x−a).

If
limx→af(x)=limx→ag(x), then the ratio
f(x)/g(x) is approximately equal to the ratio of their linear approximations near
a.

Since
f is differentiable at
a, then
f is continuous at
a, and therefore
f(a)=limx→af(x)=0. Similarly,
g(a)=limx→ag(x)=0. If we also assume that
f′ and
g′ are continuous at
x=a, then
f′(a)=limx→af′(x) and
g′(a)=limx→ag′(x). Using these ideas, we conclude that

limx→af(x)g(x)=limx→af′(x)(x−a)g′(x)(x−a)=limx→af′(x)g′(x).

Note that the assumption that
f′ and
g′ are continuous at
a and
g′(a)≠0 can be loosened. We state L’Hôpital’s rule formally for the indeterminate form
00. Also note that the notation
00 does not mean we are actually dividing zero by zero. Rather, we are using the notation
00 to represent a quotient of limits, each of which is zero.

L’hôpital’s rule (0/0 case)

Suppose
f and
g are differentiable functions over an open interval containing
a, except possibly at
a. If
limx→af(x)=0 and
limx→ag(x)=0, then

limx→af(x)g(x)=limx→af′(x)g′(x),

assuming the limit on the right exists or is
∞ or
−∞. This result also holds if we are considering one-sided limits, or if
a=∞and−∞.

You can factorize the numerator of an expression. What's the problem there? here's an example.
f(x)=((x^2)-(y^2))/2
Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y).
so the overall function becomes :
((x+y)(x-y))/2

The

The problem is the question, is not a problem where it is, but what it is

Antonio

I think you should first know the basics man: PS

Vishal

Yes, what factorization is

Antonio

Antonio bro is x=2 a function?

The

Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant

Antonio

you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though

zach

Why y, if domain its usually defined as x, bro, so you creates confusion

Antonio

Its f(x) =y=2 for every x

Antonio

Yes but he said could you put x = 2 as a function you put y = 2 as a function

zach

F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)

Antonio

x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent

zach

The said x=2 and that 2 is y

Antonio

that 2 is not y, y is a variable 2 is a constant

zach

So 2 is defined as f(x) =2

Antonio

No y its constant =2

Antonio

what variable does that function define

zach

the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2

zach

Yes true, y=2 its a constant, so a line parallel to y axix as function of y

Antonio

Sorry x=2

Antonio

And you are right, but os not a function of x, its a function of y

Antonio

As function of x is meaningless, is not a finction

Antonio

yeah you mean what I said in my first post, smh

zach

I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time

Antonio

OK you can call this "function" on a set {2}, but its a single value function, a constant

A function it's a law, that for each value in the domaon associate a single one in the codomain

Antonio

function is a something which another thing depends upon to take place.
Example
A son depends on his father.
meaning here is
the father is function of the son.
let the father be y and the son be x. the we say
F(X)=Y.

Bg

yes the son on his father

pascal

a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable